The height BM, drawn from the vertex of the corner of the rhombus ABCD, forms an angle of 30 °

The height BM, drawn from the vertex of the corner of the rhombus ABCD, forms an angle of 30 ° with the side AB, AM = 4 cm. Find the length of the diagonal BD of the rhombus if point M lies on the side AD.

If BM is the height, then triangles ABM and MBD are rectangular.

Consider triangle ABM and define BM.

tg angle ABM = AM / BM;

BM = AM / tg 30 ° = 4 / 0.5774 = 6.93 cm.

Let us define the side of the rhombus AB.

AB = AM / Sin 30 ° = 4 / 0.5 = 8 cm.

Consider triangles MBD and define BD.

In a rhombus, all sides are equal, AD = AB = 8 cm.

MD = AD – AM = 8 – 4 = 4 cm.

BD ^ 2 = MB ^ 2 + MD ^ 2;

BD ^ 2 = 6.93 ^ 2 + 42 = 48 + 16 = 64;

BD = √64 = 8 cm.

Answer: the diagonal of the rhombus BD = 8 cm.